Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $p = \dfrac{6z + 27}{3z} \div \dfrac{6(2z + 9)}{2z} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{6z + 27}{3z} \times \dfrac{2z}{6(2z + 9)} $ When multiplying fractions, we multiply the numerators and the denominators. $p = \dfrac{ (6z + 27) \times 2z } { 3z \times 6(2z + 9) } $ $ p = \dfrac {2z \times 3(2z + 9)} {3z \times 6(2z + 9)} $ $ p = \dfrac{6z(2z + 9)}{18z(2z + 9)} $ We can cancel the $2z + 9$ so long as $2z + 9 \neq 0$ Therefore $z \neq -\dfrac{9}{2}$ $p = \dfrac{6z \cancel{(2z + 9})}{18z \cancel{(2z + 9)}} = \dfrac{6z}{18z} = \dfrac{1}{3} $